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I want to prove that for any Markov Chain we have

$t_{mix}>\max_{x,A}\left(\pi(A)-\displaystyle\frac{1}{4}\right)E_x(\tau_A)$

I wanted to used that if $P_x(\tau_A\leq t)\geq \alpha $ for any $x$ in the state space then $E(\tau_x)>t_0/\alpha $. This does not work

Could anyone give a hint to solve this problem

Thanks!

1 Answers 1

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Firstly, your lemma is false. The true lemma is if there exists some $t_0>0$ and $0<\alpha<1$ such that $\mathbb{P}_x(\tau_A \leq t_0) \geq \alpha$, then $\mathbb{E}_x(\tau_A) \leq \frac{t_0}{\alpha}$.

Your idea works. Note that $\mathbb{P}_x(\tau_A \leq t_{mix}) \geq P^{t_{mix}}(x,A)$. On the other hand $\left |{P^{t_{mix}}(x,A)-\pi(A)}\right |\leq d(t_{mix})<\frac{1}{4}$, which implies $P^{t_{mix}}(x,A)>\pi(A)-\frac{1}{4}$. Then $\mathbb{P}_x(\tau_A \leq t_{mix})>\pi(A)-\frac{1}{4}$. If $\pi(A)-\frac{1}{4}\leq 0$, the inequallity that we want to prove is obvious. If $\pi(A)-\frac{1}{4}>0$, we will use your lemma to say that $\mathbb{E}_x(\tau_A) \leq \frac{t_{mix}}{\pi(A)-\frac{1}{4}}$, which implies $t_{mix} \geq \left(\pi(A)-\frac{1}{4} \right)\mathbb{E}_x(\tau_A)$, and finally we maximize the right part over all $x \in \Omega$ and $A \subseteq \Omega$, and we finish the proof.

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    To show the fact with strict inequality, only show that if $\mathbb{P}_x(\tau_A \leq t_0)>\alpha$ for some $t_0>0$ and $0<\alpha<1$, then $\mathbb{E}_x(\tau_A)<\frac { t_0} {\alpha}$, and use the fact that there exists $A \subseteq \Omega$ such that $\pi(A)-\frac {1} {4}>0$ (for example $A=\Omega$ holds this property). – 2017-02-21